<span>CH4 + 4 Cl2 → CCl4 + 4 HCl
(4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4
(4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
(2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl
11.2 mol + 3.92 mol = 15.1 mol HCl from both steps</span>
1. research question
2. background research
3. hypothesis
4. <span>Controlled experiment
5. data analysis
6. data collection
7. conclusion</span>
Answer:
Moles of
= 6 moles
Explanation:
The reaction of
and
to make
is:
⇒
The above reaction shows that 2 moles of Sc can react with 3 moles of
to form 
Mole Ratio= 2:3
For 10 moles of Sc we need:
Moles of
= 
Moles of
= 
Moles of
=15 moles
So 15 moles of
are required to react with 10 moles of
but we have 9 moles of
, it means
is limiting reactant.


Moles of ScCl_3= 6 moles
The two types of energy changes that occur are heat and light changes.
Answer: 40731.8 grams of this gasoline would fill a 14.6gal tank
Explanation:
Density is defined as the mass contained per unit volume.
Given : Mass of gasoline = ?
Density of the gasoline =
Volume of the gasoline = 14.6gal = 55267.01 ml (1gal=3785.41ml)
Putting in the values we get:
Thus 40731.8 grams of this gasoline would fill a 14.6gal tank